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Here $A$ is a commutative ring. I am confused about the problem statement. I believe that the question is referring to a generic point of $Z.$ This would correspond to a minimal prime ideal of $A$ contained in $Z.$ However, since $Z$ is irreducible, it is of the form $Z = V(\mathfrak p) :=\left\{x \in X: \mathfrak p \subseteq \mathfrak p_x \subseteq A\right\}$ for a prime ideal $\mathfrak p$ of $A.$ It seems that $\mathfrak p$ would be the unique generic point of $Z.$ My concern is that I haven't used the locally closed condition anywhere, so I suspect there is an error in my reasoning. I believe that I am somehow misunderstanding the problem statement. Could anyone shed some light on this?

EDIT -- For context, this is exercise 2.8 in Algebraic Geomtery I: Schemes, with examples and exercises by Ulrich Görtz and Torsten Wedhorn

Update: I believe that I have come up with a correct proof that I am posting below for future users to reference. Please let me know if there are errors that I haven't caught.

Proof. Let $Z$ be a nonempty (this condition is necessary, as the $\emptyset = X \cap \emptyset$ clearly doesn't contain any generic points) locally closed irreducible subset of $X.$ As $Z$ is locally closed, it is the intersection of an open subsetset $V(\mathfrak a)^c$ and a closed subsetset $V(\mathfrak b)$ of $X.$ Since $Z$ is nonempty, we have that $\mathfrak a \not\subseteq \mathfrak b.$ Hence it suffices to show that $\mathfrak b$ is prime.This would clearly make $x_\mathfrak b \in Z$ the unique generic point. To this end we note that those subsets $V$ of $Z$ which are of the form $F\cap V(\mathfrak a)^c,$ where $F\subseteq V(\mathfrak b)$ is closed are closed in $Z.$ That $Z$ is irreducible implies $V(\mathfrak b)$ is as well. As $V(\mathfrak b)$ is closed by assumption, we conclude that $\mathfrak b$ is prime.

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    $\begingroup$ Well, for one thing, the minimal prime ideals of $A$ correspond with the generic points of irreducible components of $X$, not irreducible subsets of $X$ in general. Furthermore, the prime ideals are in bijection with the irreducible closed subsets of $X$ via the correspondence $P \in Spec(A) \mapsto V(P)$, so your claim that $Z$ is of the form $V(P)$ is already assuming too much. $\endgroup$ Jan 18, 2018 at 21:52

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It is not true that every irreducible subset of $X$ has the form $V(\mathfrak{p})$. That is true only for irreducible closed subsets. (For instance, consider when $X$ is $\mathbb{A}^1$ over a field: then actually every infinite subset of $X$ is irreducible, since it has the cofinite topology, possibly with a generic point.) Since you know only that $Z$ is locally closed, you will have to do more work.

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